已知数列满足a1=0,a(n+1)=an+1+2根号(an+1),n∈N,则a99=?
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已知数列满足a1=0,a(n+1)=an+1+2根号(an+1),n∈N,则a99=?
已知数列满足a1=0,a(n+1)=an+1+2根号(an+1),n∈N,则a99=?
已知数列满足a1=0,a(n+1)=an+1+2根号(an+1),n∈N,则a99=?
a(n+1)=an+1+2(an+1),两边+1,配方,就有:
a(n+1)+1=(√(an+1)+1)^2
√(a(n+1)+1)=√(an+1)+1
即√(an+1)是公差为1的等差数列,而√(a1+1)=1
所以√(a99+1)=99
a99=99^2-1=9800
a(n+1)+1=【根号(an+1)+1】²=【根号{a(n-1)+1}+2】²=.....==【根号(a1+1)+n】²,所以a99=99²-1
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